Representing functions as power series is a key skill in AP Calculus BC, essential for financial models involving rational functions. For q(x) = x/(1+x), express it as x * 1/(1 - (-x)). Apply the geometric series $sum_{n=0}$^∞ $(-x)^n$ = 1/(1 - (-x)), then multiply by x to get $sum_{n=0}$^∞ $(-1)^n$ $x^{n+1}$. This aligns with choice B. Choice A is tempting but represents 1/(1+x) without the x multiplier, omitting the initial power shift. A transferable strategy is to factor out terms like polynomials and combine with geometric series, reindexing as needed for the correct powers.

Representing functions as power series is a key skill in AP Calculus BC, allowing us to express functions like rational ones as infinite sums for analysis. To derive the series for f(x) = 1/(1-3x), start with the geometric series $sum_{n=0}$^∞ $r^n$ = 1/(1-r) for |r|<1. Replace r with 3x, yielding $sum_{n=0}$^∞ $(3x)^n$ = $sum_{n=0}$^∞ $3^n$ $x^n$ = 1/(1-3x) for |3x|<1 or |x|<1/3. This matches choice D directly. A tempting distractor like choice C fails because it uses $(-3)^n$ $x^n$, which would represent 1/(1+3x) instead of 1/(1-3x), altering the sign and thus the function. Remember, a transferable strategy is to manipulate known series like the geometric one by substitution and algebraic adjustments to match the target function.

This question tests the skill of representing functions as power series, using the cosine expansion. The Maclaurin series for cos(y) is $\sum_{n=0}^{\infty} (-1)^n \frac{y^{2n}}{(2n)!}$ for all real y. Replacing y with 2x gives $\sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n} x^{2n}}{(2n)!}$, but choice A keeps it as $(2x)^{2n}$. This is equivalent and converges everywhere. A tempting distractor is choice B, which has odd powers and represents sin(2x) instead. A transferable strategy for representing functions as power series is to substitute into standard Taylor series for transcendental functions like cosine.

This problem asks for the power series of r(x) = 2/(1-4x). We factor out the constant: r(x) = 2·1/(1-4x), where the geometric series gives 1/(1-4x) = Σ(n=0 to ∞) $(4x)^n$ = Σ(n=0 to ∞) $4^n$ $x^n$. Therefore, r(x) = 2·Σ(n=0 to ∞) $4^n$ $x^n$ = Σ(n=0 to ∞) $2·4^n$ $x^n$. Choice A would have 2·4^(n+1) = $2·4·4^n$ = $8·4^n$ as coefficients, which is four times too large. When constants appear in the numerator, factor them out and multiply the entire series by that constant.

This problem requires the power series for u(x) = e^(2x). The exponential series is $e^y$ = Σ(n=0 to ∞) $y^n$/n! for all y. With y = 2x, we get u(x) = Σ(n=0 to ∞) $(2x)^n$/n! = Σ(n=0 to ∞) $2^n$ $x^n$/n!. Choice A incorrectly separates the 2 from $x^n$, giving only $2x^n$ instead of $(2x)^n$ = $2^n$ $x^n$ in the numerator. The key is to substitute the entire expression 2x into the exponential series, keeping 2x grouped together when raising to the nth power.

This question tests the skill of representing functions as power series, using the exponential series. The Maclaurin series for $e^y$ is $ \sum_{n=0}^{\infty} \frac{y^n}{n!} $ for all real y. Substituting y = 2x gives $ \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!} $, which is the series for e^{2x}. This holds for all real x since the exponential series converges everywhere. A tempting distractor is choice A, which has (n+1)! in the denominator, altering it to something like an integral of the exponential. A transferable strategy for representing functions as power series is to substitute scaled variables into known Taylor series like the exponential.

Representing functions as power series is a key skill in AP Calculus BC, applied in models like displacement with inverse trig functions. The series for arctan(x) comes from integrating the geometric series for $1/(1+t^2$) = $sum_{n=0}$^∞ $(-1)^n$ $t^{2n}$ for |t|<1. Integrating from 0 to x yields arctan(x) = $sum_{n=0}$^∞ $(-1)^n$ integral $t^{2n}$ dt = $sum_{n=0}$^∞ $(-1)^n$ $x^{2n+1}$/(2n+1). This is choice B. Choice D fails as a distractor by using $(-1)^{n+1}$, which negates the series and represents -arctan(x). A transferable strategy is to integrate known geometric series adaptations for inverse functions, carefully handling the powers and signs.

Representing functions as power series is a key skill in AP Calculus BC, enabling approximation and integration of functions like logarithms. The series for ln(1+x) derives from integrating the geometric series for 1/(1+t) = $sum_{n=0}$^∞ $(-1)^n$ $t^n$ for |t|<1. Integrating from 0 to x gives ln(1+x) = integral $sum_{n=0}$^∞ $(-1)^n$ $t^n$ dt = $sum_{n=0}$^∞ $(-1)^n$ $x^{n+1}$/(n+1), then reindexing to n=1 yields $sum_{n=1}$^∞ $(-1)^{n-1}$ $x^n$ / n. This corresponds to choice A. Choice C is a common distractor but fails as it has $(-1)^n$ instead of $(-1)^{n-1}$, flipping the signs and representing -ln(1+x). A transferable strategy is to derive new series by integrating or differentiating known ones, adjusting indices accordingly for the desired function.

This problem asks for the power series of $g(x) = \frac{5}{1+2x}$. We can rewrite this as $5 \cdot \frac{1}{1 - (-2x)}$, matching the geometric series form $\frac{1}{1-r} = \sum r^n$. With $r = -2x$, we get $g(x) = 5 \cdot \sum_{n=0}^{\infty} (-2x)^n = 5 \cdot \sum_{n=0}^{\infty} (-2)^n x^n = \sum_{n=0}^{\infty} 5(-2)^n x^n$. Choice A incorrectly writes $(2x)^n$ instead of $(-2x)^n$, missing the alternating signs from the $+2x$ in the denominator. When the denominator has $1 + ax$, remember to use $r = -ax$ in the geometric series.

This problem asks for the power series representation of g(x) = 1/(4+x) about x = 0. We first factor out 4 from the denominator to get g(x) = 1/4 · 1/(1+x/4). Using the geometric series 1/(1-u) = Σ(n=0 to ∞) $u^n$ with u = -x/4, we obtain 1/(1-(-x/4)) = Σ(n=0 to ∞) $(-x/4)^n$. Therefore, g(x) = 1/4 · Σ(n=0 to ∞) $(-1)^n$ $(x/4)^n$ = Σ(n=0 to ∞) $(-1)^n$ $x^n$/4^(n+1), which converges for |x/4| < 1 or |x| < 4. Choice A incorrectly places the factor of 1/4 inside the sum without adjusting the exponent. When factoring constants from denominators, always factor completely before applying the geometric series formula.